B
anachJ
ournal ofM
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nalysis ISSN: 1735-8787 (electronic)http://www.math-analysis.org
REVERSE OF THE GRAND FURUTA INEQUALITY AND ITS APPLICATIONS
MASATOSHI FUJII1∗, RITSUO NAKAMOTO2AND MASARU TOMINAGA3 This paper is dedicated to Professor J.E. Peˇcari´c
Submitted by A. R. Villena
Abstract. We shall give a norm inequality equivalent to the grand Furuta inequality, and moreover show its reverse as follows: LetAandB be positive operators such that 0 < m ≤ B ≤ M for some scalars 0 < m < M and h:= Mm >1. Then
kA12{A−2t(Ar2B(r−t){(p−t)s+r}
1−t+r Ar2)1sA−t2}1pA12 k
≤ K(hr−t,(p−t)s+r
1−t+r )ps1 kA1−t+r2 Br−tA1−t+r2 kps(1−t+r)(p−t)s+r
for 0≤ t≤ 1,p≥ 1,s ≥1 and r ≥t ≥0, where K(h, p) is the generalized Kantorovich constant. As applications, we consider reverses related to the Ando-Hiai inequality.
1. Introduction
The origin of reverse inequalities is the Kantorovich inequality. It says that if a positive operator A on a Hilbert space H satisfies 0 ≤m≤A≤M, then
hA−1x, xi ≤ (M +m)2
4M m hAx, xi−1 for all unit vectorsx∈H. (K)
Date: Received: 15 April 2008; Accepted: 20 April 2008.
∗ Corresponding author.
2000Mathematics Subject Classification. 47A63.
Key words and phrases. grand Furuta inequality, Furuta inequality, L¨owner-Heinz inequal- ity, Araki-Cordes inequality, Bebiano-Lemos-Providˆencia inequality, norm inequality, positive operator, operator inequality, reverse inequality.
23
The point in (K) is the convexity of the function t → t−1. Mond and Peˇcari´c turned their attention to the convexity of functions, and established the so called Mond-Peˇcari´c method in the theory of reverse inequalities, see [13] in detail. The subject of this note is just on the line of Mond-Peˇcari´c’s idea, and our target is the grand Furuta inequality.
LetAand B be positive (bounded linear) operators acting on a Hilbert space.
The grand Furuta inequality [10] says that
A≥B ≥0 ⇒ A1−t+r ≥ {Ar2(A−t2BpA−2t)sAr2}(p−t)s+r1−t+r (GFI) for 0≤t ≤1,p≥1,s ≥1 and r ≥t.
The inequality (GFI) is considered as a parametric formula interpolating the Furuta inequality (FI) and Ando-Hiai one (1.1), respectively [9] and [1]:
A≥B ≥0 ⇒ A1+r≥(Ar2BpAr2)1+rp+r (r≥0, p≥1) (FI) and
A≥B ≥0 ⇒ Ar ≥ {Ar2(A−12BpA−12)rAr2}1p (p, r≥1). (1.1) Now the Furuta inequality appeared as a useful extension of the so-called L¨owner-Heinz inequality (cf. [14]):
A≥B ≥0 ⇒ Aα ≥Bα (0≤α≤1). (1.2) This L¨owner-Heinz inequality (1.2) is equivalent to the Araki-Cordes inequality ([2], [4]):
kAp2BpAp2 k ≤ kA12BA12 kp (0≤p≤1). (1.3) M.Fujii and Y.Seo [8] gave a reverse inequality of the Araki-Cordes inequality:
If A and B are positive operators such that 0 < m ≤ B ≤ M for some scalars 0< m < M and h:= Mm (>1), then
K(h, p) kA12BA12 kp ≤ kAp2BpAp2 k (0≤p≤1) (1.4) where a generalized Kantorovich constantK(h, p) is defined as follows:
K(h, p) := 1 h−1
hp−h p−1
p−1
hp−h
hp−1 p
p
(1.5) for all h(6= 1), p∈R and K(h,0) =K(h,1) = 1, see [11] and [13].
In this note, we first give a norm inequality equivalent to the grand Furuta inequality (GFI). Based on this, we show a reverse inequality of (GFI), in which the generalized Kantorovich constant (1.5) is used. As an application, we obtain reverses of a generalization of Ando-Hiai inequality (1.1).
2. Norm Inequality equivalent to the grand Furuta inequality The grand Furuta inequality (GFI) is equivalent to the following norm inequal- ity:
Lemma 2.1. LetAandB be positive operators. Then the grand Furuta inequality (GF I) is equivalent to
kA1−t+r2 Br−tA1−t+r2 kps(1−t+r)(p−t)s+r ≤ k A12{A−2t(Ar2B(r−t){(p−t)s+r}
1−t+r Ar2)1sA−2t}1pA12 k (2.1) for 0≤t≤1, p≥1, s ≥1 and r≥t.
Proof. Replace A to A−1 and put
C ={A2t(A−r2B(r−t){(p−t)s+r}
1−t+r A−r2)1sA2t}1p in (2.1). Since Br−t={Ar2(A−t2CpA−t2)sAr2}(p−t)s+r1−t+r , we have
kA−1−t+r2 {Ar2(A−2tCpA−2t)sAr2}(p−t)s+r1−t+r A−1−t+r2 kps(1−t+r)(p−t)s+r ≤ kA−12CA−12 k. This is equivalent to the inequality
A≥C ⇒ A1−t+r ≥ {Ar2(A−t2CpA−t2)sAr2}(p−t)s+r1−t+r ,
that is, (2.1) is equivalent to the grand Furuta inequality (GFI).
Corollary 2.2. Let A and B be positive operators. Then
kA1+s2 B1+sA1+s2 kp(1+s)p+s ≤ kA12(As2Bp+sAs2)p1A12 k (2.2) for p≥1 and s≥0.
Moreover
kA1+t2 BtA1+t2 k ≤ k A12(A2sBsAs2)tsA12 k (2.3) for s≥t≥0.
Proof. Put t = 0, s = 1 in (2.1). Then replacing r and B to s and B1+ss , respectively, (2.1) implies (2.2).
Moreover, let t be a real number satisfying s≥t ≥0. Then (2.2) implies kA1+t2 B1+tA1+t2 kp(1+t)p+s ≤ kA1+s2 B1+sA1+s2 kp(1+s)p+s ≤ kA12(As2Bp+sAs2)1pA12 k by 1+s1+t ∈[0,1] and the Araki-Cordes inequality (1.3). Furthermore, replacing B
toB1+tt and puttingp= st, we have (2.3).
Remark 2.3. The inequality (2.3) is originated by Bebiano-Lemos-Providˆencia in [3]. In our previous note [7], we call it the BLP inequality and we showed (2.2) as a generalization of the BLP inequality (2.3). Incidentally it is equivalent to (F I). For convenience, we give a proof of (2.2) ⇒ (F I). The inequality (2.2) is rephrased by replacingA to A−1 as follows:
kA−1+t2 BtA−1+t2 kp(1+t)p+s ≤ kA−12(A−s2Bt(p+s)1+t A−s2)1pA−12 k. Moreover, putting
C = (A−s2Bt(p+s)1+t A−s2)1p, orBt= (As2CpAs2)p+s1+t,
it is also rephrased as
kA−1+t2 (As2CpAs2)p+s1+tA−1+t2 kp(1+t)p+s ≤ kA−12CA−12 k which obviously implies the Furuta inequality (F I) by taking s=t=r.
Remark 2.4. In [12], Furuta gave a similar inequality to (2.1).
3. A reverse grand Furuta inequality and its applications In this section, we give a reverse inequality of (2.1) by using the generalized Kantorovich constant (1.5).
Theorem 3.1. Let A and B be positive operators such that0< m≤B ≤M for some scalars 0< m < M and h:= Mm >1. Then
kA12{A−2t(Ar2B(r−t){(p−t)s+r}
1−t+r Ar2)1sA−2t}1pA12 k
≤ K
h1−t+r
0 1−t+r(r−t)
,(p−t)s+r 1−t+r0
ps1
kA1−t+r
0
2 B1−t+r
0 1−t+r(r−t)
A1−t+r
0
2 k
(p−t)s+r ps(1−t+r0)
(3.1) for0≤t ≤1, p≥1, s≥1and1+r ≥1+r0 > t, whereK(h, p)is the generalized Kantorovich constant defined by (1.5).
Proof. For p≥1 ands ≥1, the Araki-Cordes inequality (1.3) implies that kA12{A−t2(Ar2B(r−t){(p−t)s+r}
1−t+r Ar2)1sA−t2}p1A12 k
≤ kAp2{A−2t(Ar2B(r−t){(p−t)s+r}
1−t+r Ar2)1sA−t2}Ap2 k1p
= kAp−t2 (Ar2B(r−t){(p−t)s+r}
1−t+r Ar2)1sAp−t2 k1p
≤ kA(p−t)s2 (Ar2B(r−t){(p−t)s+r}
1−t+r Ar2)A(p−t)s2 kps1
= kA(p−t)s+r2 B(r−t){(p−t)s+r}
1−t+r A(p−t)s+r2 kps1 .
Moreover, since (p−t)s+r≥1−t+r0 >0, it follows from the reverse Araki-Cordes inequality (1.4) that
kA(p−t)s+r2 B(r−t){(p−t)s+r}
1−t+r A(p−t)s+r2 kps1
≤ kA(p−t)s+r2 B(r−t)1−t+r
0 1−t+r
(p−t)s+r
1−t+r0 A(p−t)s+r2 kps1
≤ K
h1−t+r
0 1−t+r(r−t)
,(p−t)s+r 1−t+r0
ps1
kA1−t+r
0
2 B1−t+r
0 1−t+r(r−t)
A1−t+r
0
2 k
(p−t)s+r ps(1−t+r0) .
Combining them, we have the desired inequality (3.1).
From the reverse grand Furuta inequality (3.1) we have the following reverse Furuta inequality (see [7]):
Corollary 3.2. Let A and B be positive operators such that 0 < m ≤ B ≤ M for some scalars 0< m < M and h:= Mm >1. Then
kA12(As2Bp+sAs2)1pA12 k ≤ K
h1+t,p+s 1 +t
1p
kA1+t2 B1+tA1+t2 kp(1+t)p+s (3.2) for all p≥1 and s≥t >−1.
Proof. In (3.1), if we put t = 0, s = 1, and replace r, r0, B and h to s, t, B1+ss and h1+ss , respectively, then the desired inequality (3.2) holds.
On the other hand, Ando and Hiai [1] proved
A]αB ≤1 ⇒ Ar]αBr ≤1 for 0≤α≤1, r≥1 where A]αB :=A12(A−12BA−12)αA12. This inequality is equivalent to
kAr]αBr k ≤ kA]αB kr . (AH) M.Fujii and E.Kamei [6] proved that (AH) is equivalent to (FI). Also they extended (AH) as follows:
kAr] αr
(1−α)s+αrBs k(1−α)s+αrsr ≤ kA]αB k (GAH)
for r, s≥ 1 and 0 ≤α ≤ 1. It is easy to see that the inequality (2.1) equivalent to the grand Furuta inequality is rewritten as follows:
kA1−t+r2 (A−r2BsA−r2)(p−t)s+r1−t+r A1−t+r2 kps(1−t+r)(p−t)s+r ≤ kA12(A−2tBA−2t)1pA12 k for 0≤t ≤1,p≥1,s ≥1 and r ≥t≥0. Here if we put α= 1p, then we have
kA1−t+r2 (A−r2BsA−r2)
α(1−t+r)
(1−αt)s+αrA1−t+r2 k
(1−αt)s+αr
s(1−t+r) ≤ kA12(A−2tBA−2t)αA12 k. (3.3) This inequality (3.3) implies (GAH) by t= 1.
From the viewpoint of the Ando-Hiai inequality, we consider the following inequality related to a reverse inequality of (3.3) which is equivalent to (3.1).
Theorem 3.3. Let A and B be positive operators such that 0< m≤A, B ≤M for some scalars 0< m < M and h:= Mm >1. Then
K
hr+s, α(1−t+r0) (1−αt)s+αr
kA12(A−2tBA−2t)αA12 k s(1−t+r
0) (1−αr)s+αr
≤ kA1−t+r
0
2 (A−r2BsA−r2)
α(1−t+r0 )
(1−αt)s+αrA1−t+r
0
2 k
(3.4)
for 0 ≤ t ≤ 1, s ≥ 1, 1 +r ≥ 1 +r0 ≥ t and 0 ≤ α ≤ 1 where K(h, p) is the generalized Kantorovich constant defined by (1.5).
Proof. In (3.1), we replaceBr−t,hr−tandpto (A−r2BsA−r2)
α(1−t+r) (1−αt)s+αr,h
α(r+s)(1−t+r) (1−αt)s+αr
and α1, respectively. Then we have kA12(A−t2BA−t2)αA12 k ≤ K
h
α(r+s)(1−t+r0)
(1−αt)s+αr ,(1−αt)s+αr α(1−t+r0)
αs
× kA1−t+r
0
2 (A−r2BsA−r2)
α(1−t+r0)
(1−αt)s+αrA1−t+r
0
2 k(1−αt)s+αrs(1−t+r0) . By the inversion formula (i.e., K(hr,1r) =K(h, r)−1r for all r6= 0) [5], it implies
K
h
α(r+s)(1−t+r0 )
(1−αt)s+αr ,(1−αt)s+αr α(1−t+r0)
αs
= K
hr+s, α(1−t+r0) (1−αt)s+αr
−(1−αt)s+αrs(1−t+r0
)
,
and hence (3.4) holds.
Remark 3.4. If r=r0 in (3.4), then we have the following reverse inequality of (3.3):
K
hr+s, α(1−t+r) (1−αt)s+αr
kA12(A−2tBA−2t)αA12 k
s(1−t+r) (1−αr)s+αr
≤ kA1−t+r2 (A−r2BsA−r2)
α(1−t+r)
(1−αt)s+αrA1−t+r2 k
for 0 ≤t ≤ 1, s ≥ 1, 1 +r ≥ t and 0 ≤ α ≤ 1. Moreover, let t = 1 in Theorem 3.3. As a reverse inequality of (GAH), we have
K
hr+s, αr (1−α)s+αr
kA12(A−12BA−12)αA12 k(1−α)s+αrsr
≤ kAr2(A−r2BsA−r2)(1−α)s+αrαr Ar2 k, that is,
K
hr+s, αr (1−α)s+αr
kA]αB k(1−α)s+αrsr ≤ kAr] αr
(1−α)s+αrBsk for s≥1,r ≥0 and 0≤α≤1.
Under the conditions of 0≤s≤1 andr0 =r, we prove the following inequality as in Theorem 3.3:
Theorem 3.5. Let A andB be positive operators on a Hilbert space H such that 0< m≤A, B ≤M for some scalars 0< m < M and h:= Mm >1. Then
kA1−t+r2 (A−r2BsA−r2)
α(1−t+r)
(1−αt)s+αrA1−t+r2 k
≤ K(h1+t, α)−
s(1−t+r)
(1−αt)s+αr kA12(A−t2BA−2t)αA12 k(1−αt)s+αrs(1−t+r)
(3.5) for0≤s, t≤1, 1 +r≥t and 0≤α≤1with α(1−t)≤(1−αt)s where K(h, p) is the generalized Kantorovich constant defined by (1.5).
Proof. We use the H¨older-McCarthy inequality and its reverse: LetAbe a positive operator with 0< m≤A≤M. Then for every vectory ∈H
K(h, β)hAy, yiβ kyk2(1−β) ≤ hAβy, yi ≤ hAy, yiβ kyk2(1−β) for 0≤β ≤1.
Since Mmt ≤ mA−t ≤ A−t2BA−t2 ≤M A−t ≤ mMt and kAγxk ≤ k Aγ k = kA kγ
≤ Mγ for all unit vectors x∈H and γ >0, we have for any 0≤s≤1 hA1−t+r2 (A−r2BsA−r2)
α(1−t+r)
(1−αt)s+αrA1−t+r2 x, xi
≤ hA1−t2 BsA1−t2 x, xi(1−αt)s+αrα(1−t+r) kA1−t+r2 xk2{1−(1−αt)s+αrα(1−t+r) }
≤ hA1−t2 BA1−t2 x, xi(1−αt)s+αrsα(1−t+r) kA1−t2 xk2(1−s)α(1−t+r)
(1−αt)s+αr M(1−αt)s+αr1−t+r (s−αst−α+αt)
≤ (K(h1+t, α)−1hA12(A−t2BA−t2)αA12x, xi)(1−αt)s+αrs(1−t+r) kA12xk−2(1−α)s(1−t+r) (1−αt)s+αr
× M(1−αt)s+αr1−t+r (α(1−s)(1−t))
M(1−αt)s+αr1−t+r (s−αst−α+αt)
≤ K(h1+t, α)−
s(1−t+r)
(1−αt)s+αr kA12(A−t2BA−t2)αA12 k
s(1−t+r) (1−αt)s+αr
× M−(1−αt)s+αr1−t+r (1−α)sM(1−αt)s+αr1−t+r (s−αs)
= K(h1+t, α)−
s(1−t+r)
(1−αt)s+αr kA12(A−2tBA−2t)αA12 k(1−αt)s+αrs(1−t+r) .
Hence we obtain the desired inequality (3.5).
Puttingt = 1 in (3.5), we have an inequality given in [15]:
kAr] αr
(1−α)s+αrBs k ≤ K(h2, α)−(1−α)s+αrrs kA]αB k(1−α)s+αrrs for 0≤s ≤1, r≥0 and 0≤α≤1.
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1,2Faculty of Engineering, Ibaraki University, Hitachi, Ibaraki 316-0033, Japan.
E-mail address: 1[email protected], 2[email protected]
3 Toyama National College of Technology, Hongo-machi, Toyama-shi 939- 8630, Japan.
E-mail address: [email protected]
